Bijection iff Left and Right Inverse/Corollary

Corollary to Bijection iff Left and Right Inverse

Let $f: S \to T$ and $g: T \to S$ be mappings such that:

\(\ds g \circ f\)

\(=\)

\(\ds I_S\)

\(\ds f \circ g\)

\(=\)

\(\ds I_T\)

Then both $f$ and $g$ are bijections.

Proof

Suppose we have such mappings $f$ and $g$ with the given properties.

From Bijection iff Left and Right Inverse, we have that $f$ is a bijection, by considering $g = g_1$ and $g = g_2$.

It directly follows that by setting $g = f, f = g_1, f = g_2$, the result Bijection iff Left and Right Inverse can be used the other way about.

$\blacksquare$

Sources

• 1951: Nathan Jacobson: Lectures in Abstract Algebra: Volume $\text { I }$: Basic Concepts ... (previous) ... (next): Introduction $\S 2$: Product sets, mappings
• 1978: Thomas A. Whitelaw: An Introduction to Abstract Algebra ... (previous) ... (next): $\S 25.1$: Some further results and examples on mappings